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| 1 | +# Binary Tree Level Order Traversal II |
| 2 | + |
| 3 | +## Question |
| 4 | + |
| 5 | +- leetcode: [Binary Tree Level Order Traversal II | LeetCode OJ](https://leetcode.com/problems/binary-tree-level-order-traversal-ii/) |
| 6 | +- lintcode: [(70) Binary Tree Level Order Traversal II](http://www.lintcode.com/en/problem/binary-tree-level-order-traversal-ii/) |
| 7 | + |
| 8 | +``` |
| 9 | +Given a binary tree, return the bottom-up level order traversal of its nodes' values. |
| 10 | +(ie, from left to right, level by level from leaf to root). |
| 11 | +
|
| 12 | +Example |
| 13 | +Given binary tree {3,9,20,#,#,15,7}, |
| 14 | +
|
| 15 | + 3 |
| 16 | + / \ |
| 17 | + 9 20 |
| 18 | + / \ |
| 19 | + 15 7 |
| 20 | +
|
| 21 | +
|
| 22 | +return its bottom-up level order traversal as: |
| 23 | +[ |
| 24 | + [15,7], |
| 25 | + [9,20], |
| 26 | + [3] |
| 27 | +] |
| 28 | +``` |
| 29 | + |
| 30 | +## 題解 |
| 31 | + |
| 32 | +這題在普通的 BFS 基礎上增加了逆序輸出,簡單的實現可以使用 Stack 或者最後對結果逆序。 |
| 33 | + |
| 34 | +### Java - Stack |
| 35 | + |
| 36 | +```java |
| 37 | +/** |
| 38 | + * Definition of TreeNode: |
| 39 | + * public class TreeNode { |
| 40 | + * public int val; |
| 41 | + * public TreeNode left, right; |
| 42 | + * public TreeNode(int val) { |
| 43 | + * this.val = val; |
| 44 | + * this.left = this.right = null; |
| 45 | + * } |
| 46 | + * } |
| 47 | + */ |
| 48 | + |
| 49 | + |
| 50 | +public class Solution { |
| 51 | + /** |
| 52 | + * @param root: The root of binary tree. |
| 53 | + * @return: buttom-up level order a list of lists of integer |
| 54 | + */ |
| 55 | + public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) { |
| 56 | + ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); |
| 57 | + if (root == null) return result; |
| 58 | + |
| 59 | + Stack<ArrayList<Integer>> s = new Stack<ArrayList<Integer>>(); |
| 60 | + Queue<TreeNode> q = new LinkedList<TreeNode>(); |
| 61 | + q.offer(root); |
| 62 | + while (!q.isEmpty()) { |
| 63 | + int qLen = q.size(); |
| 64 | + ArrayList<Integer> aList = new ArrayList<Integer>(); |
| 65 | + for (int i = 0; i < qLen; i++) { |
| 66 | + TreeNode node = q.poll(); |
| 67 | + aList.add(node.val); |
| 68 | + if (node.left != null) q.offer(node.left); |
| 69 | + if (node.right != null) q.offer(node.right); |
| 70 | + } |
| 71 | + s.push(aList); |
| 72 | + } |
| 73 | + |
| 74 | + while (!s.empty()) { |
| 75 | + result.add(s.pop()); |
| 76 | + } |
| 77 | + return result; |
| 78 | + } |
| 79 | +} |
| 80 | +``` |
| 81 | + |
| 82 | +### Java - Reverse |
| 83 | + |
| 84 | +```java |
| 85 | +/** |
| 86 | + * Definition of TreeNode: |
| 87 | + * public class TreeNode { |
| 88 | + * public int val; |
| 89 | + * public TreeNode left, right; |
| 90 | + * public TreeNode(int val) { |
| 91 | + * this.val = val; |
| 92 | + * this.left = this.right = null; |
| 93 | + * } |
| 94 | + * } |
| 95 | + */ |
| 96 | + |
| 97 | + |
| 98 | +public class Solution { |
| 99 | + /** |
| 100 | + * @param root: The root of binary tree. |
| 101 | + * @return: buttom-up level order a list of lists of integer |
| 102 | + */ |
| 103 | + public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) { |
| 104 | + ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); |
| 105 | + if (root == null) return result; |
| 106 | + |
| 107 | + Queue<TreeNode> q = new LinkedList<TreeNode>(); |
| 108 | + q.offer(root); |
| 109 | + while (!q.isEmpty()) { |
| 110 | + int qLen = q.size(); |
| 111 | + ArrayList<Integer> aList = new ArrayList<Integer>(); |
| 112 | + for (int i = 0; i < qLen; i++) { |
| 113 | + TreeNode node = q.poll(); |
| 114 | + aList.add(node.val); |
| 115 | + if (node.left != null) q.offer(node.left); |
| 116 | + if (node.right != null) q.offer(node.right); |
| 117 | + } |
| 118 | + result.add(aList); |
| 119 | + } |
| 120 | + |
| 121 | + Collections.reverse(result); |
| 122 | + return result; |
| 123 | + } |
| 124 | +} |
| 125 | +``` |
| 126 | + |
| 127 | +### 源碼分析 |
| 128 | + |
| 129 | +Java 中 Queue 是接口,通常可用 LinkedList 實例化。 |
| 130 | + |
| 131 | +### 複雜度分析 |
| 132 | + |
| 133 | +時間複雜度爲 $$O(n)$$, 使用了 Queue 或者 Stack 作爲輔助空間,空間複雜度爲 $$O(n)$$. |
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